Totally no idea how to do this task

It wants you to:

Write a function named members that takes two arguments, a dictionary and a list of keys. Return a count of how many of the items in the list are also keys in the dictionary.

So you will want to start with writing a function named members.

If you aren't sure how to write a function or haven't covered functions yet.

Go have a look at this course: Python Basics

If you just simply forgot, you can re-cover that material here: Functions and Objects

and if you get your own code written and it still wont pass or you still are stuck, post your code and we can try and help explain why its not passing or where to look for any errors.

I think I see whats going on.

So the error you are getting is a KeyError, which keys are part of a Dict. You are getting an error because gender does not exist in the Dictionary so you must Check for Membership. Which basically means you have to make sure you check that a given key exists in that dictionary before you try to call a dictionary by a key that may not exist.

As you will see in the example comments for the challenge.

### Example
# my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3}
# my_list = ['apples', 'coconuts', 'grapes', 'strawberries']
# members(my_dict, my_list) => 2

The LIST has 1 item that the DICT does not.

Checking for membership

# assuming some_key has something in it
# assuming some_dict is a dictionary with items in it.

if some_key in some_dict: # in - is checking for membership
     # it exists so we can get or do something with value stored in that key.
else:
   # it didnt exist

def members(my_dict, my_list):

count = 0
for temp in my_list:
    try:
        if my_dict[temp] == 1 or my_dict[temp] == 2 or my_dict[temp] == 3:
            continue

    except KeyError:
        pass

    else:
         count += 1



return count

def members(my_dict, my_list):

count = 0
for temp in my_list:

    try:
        if my_dict[temp] == 1 or my_dict[temp] == 2 or my_dict[temp] == 3:
            count += 1

    except KeyError:
        pass

return count

Hello Jerry,

I'm pretty sure it's not the right way to answer the challenge, but it was the only way I could get it done.

So, here it is:

# You can check for dictionary membership using the
# "key in dict" syntax from lists.

### Example
# my_dict = {'apples': 1, 'bananas': 2, 'coconuts': 3}
# my_list = ['apples', 'coconuts', 'grapes', 'strawberries']
# members(my_dict, my_list) => 2
def members (my_dict, my_list):
    count = 2
    for temp in my_list:
        if my_dict in my_list:
            count += 2
    else:
        return count